SHM SOLUTION 1,2,3,4,5 - Solutions.pdf

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    IN CHAPTER EXERCISE 1

    SOLUTION

    1. 0.5 s

    5 5sin t, sin t 1 sin2

    or 1

    t s2 = 0.5 s

    2. x = 4sin10πt

    amplitude = 4 cm; frequency , v = 5 Hz

    angular frequency, 2 v 10 rad s-1

    At t = 0, 0 asin   or 0

    Use x a sin( t )

    3. (i) 8 cm s-2 (ii) 4 cm s-2

    (i) acceleration =2

    2

    2

    4A A

    T

    =2

    2

    2

    42cms

    = 8 cm s-2

    (ii) acceleration =

    2

    224x xT

    =2

    2

    2

    41cms

    = 4 cm s-2

    4. (a) 0.02 m (b) 4s

    (c) 3.142 × 10-2 m s-1 (d) 4.94 × 10-2 ms-2

    Comparing with 0x Asin( t ) , we get(a) A= 0.02 m

    (b)2

    0.5 ;2 T 2

    or T = 4s

    (c)1

    maxv A 0.02 ms2

     = 0.01× 3.142 ms-1 = 3.142 × 10-2 ms-1

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    (d)2

    2 2maxa A 0.02ms

    4  =

    2484 0.02 ms49 4

     = 4.94 × 10-2 ms-2

    IN CHAPTER EXERCISE 2

    SOLUTION

    1. (i) 4.4 × 10-5 J (ii) 3.3 × 10-5 J (iii) 1.1 × 10-5 J

    Total energy =2

    2 2 2 21 1 4m A 0.2 (2 10 ) J2 2 36

    = 4.4 × 10-5 J

    Kinetic energy =2 2 21 m (a x )

    2  =

    24 41 40.2 [4 10 1 10 ]J

    2 36

    = 3.3 × 10-5 J

    Potential energy = (4.4 × 10-5 – 3.3 × 10-5) J

    = 1.1 × 10-5

     J

    2. (i)A

    2(ii)

    3± A

    2

    (i)2 2 21 1k(A x ) kx

    2 2

    (ii) when max1

    v v2

    2max

    1 1K.E. (K.E.) kA4 8

    2 2 21 1k(A x ) kA

    2 8

    3. (a) 0.314 ms-1 (b) 0.1 J (c) 0.1 J (d) 0.083 J

    (a)2 1

    max

    22v A 2.5 10 2 2 ms

    7  = 0.314ms-1

    (b)2 2 2

    max1 1E m A mv2 2

     = 1 2 0.314 J2

     = 0.1 J

    (c) Maximum potential energy = 0.1 J

    (d) Kinetic energy =2 2 21 m (A x )

    2  =

    2

    2 2 2 21 222 2 2 [(2.5 10 ) (1 10 ) ]2 7

    = 0.083 J

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    4. (a) 2 2v A x

    or  2 2 2 2v (A x )

     Now, 2 2 2 20.03 (A 0.04 )

    and  2 2 2 20.04 (A 0.03 )

    On simplification, A = 0.05 m and  = rad s-1

    Time period,2

    T 2 s

     = 2 × 3.142 s = 6.284 s

    (b) Energy =2 21 mA

    2

    =31 50 10 0.05 0.05 1 1J

    2

    = 6.25 × 10-5 J

    IN CHAPTER EXERCISE 3

    SOLUTION

    1. 8.5 s

    LT 2

    g ,

    2 1

    1 2

    T g

    T g

    Required time period =10

    3.4

    1.6

    2. The time period is independent of the mass of bob.

    3. Due to electric force of attraction between the bob and the plate, the effective value of g shall

    increase. Since T 2g

      l

     therefore T shall decrease.

    4. 0.16 ms-1

    v A  = 0.05 × 22

    5. -1 2 -20.02π ms , 0.02π ms

    Time period = time taken in one oscillation = 2 s

    v A 22

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    2

    2 2 2a A100 2

    IN CHAPTER EXERCISE 4

    SOLUTION

    1. 3.33 rad s-1

    k m

    2. 1 2

    1 2

    m(k + k )2π

    k k

    The equivalent force constant is1 2

    1 2

    k k 

    k k 

    3. 0.54 s

    Effective force constant = 40 Nm-1

    Time period,0.3

    T 2 s40

    = 0.54 s

    4. (a) 0.31 s (b) 20 ms-2 (c) 1.0 ms-1

    (a)m

    T 2k 

    (b) max.k 

    acc. Am

    (c) max. velocity =k 

    Am

    5. 1.1 × 102

     N m-1

    , 36 kg2

    2

    m 2 mT 2 ; k  

    2k T

    Here m = 12 kg and T = 1.5 s

    After the block has been placed on the tray, mass is (M + 12)kg.

     Now, M 12T 22k 

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    EXERCISE 1

    1. A 2. C 3. D 4. A 5. D 6. D

    7. C 8. D 9. C 10. C 11. D 12. D 13

    C 14. B 15. C 16. B 17. A 18. C 19. B

    20. C 21. B 22. D 23. A 24. B 25. A

    26. C 27. C 28. B 29. (B) 30. (A) 31. (D)32. (C) 33. (B) 34. (C) 35. (B) 36. (C) 37. (B)

    38. (A) 39. (A) 40. (A) 41. (C) 42. (C) 43. (B)

    44. (B) 45. (D)

    SOLUTION

    1. (A)

    v A 2 cos 2 t 3

    max.v v cos 2 t 13

    2 t3

    t = 1/3

    2. (C)

    2

    a A v

    v AA

    2

    2

    va A

    A

    2va

    A 'A' doubled  'a' halved 

    3. (D)

    P.E.min at mean position = 5 JT.E. = 9 Jmax K.E. = 4 J

    2 21 mA 4J2

     200

    2T sec.

    200 100

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    4. (A)

    max. acceleration of plank should not exceed g2A g

    2A 10 / a = g (when block leaves contact)

    5. (D)

    Particle starts from mean position.

    x Asin( t) at t = 1

    1

    2x Asin 1

    8

    1

    Ax

    2

    at t = 2

    x2 = A

    distance covered in 1st second =A

    2

    distance covered in 2nd  secondA

    A2

    ratio1/ 2 1

    2 11 2 112

    6. (D)

    Circular representation at t = 0Phase difference = 2 / 3Phase covered by each particle =  / 3

    Time takenT T

    60360 6

    7. (C)

    For max distance Vres. = 0

    v1 = v2 x1 = x2x1 + x2 = 20 cmx1 = 10 cm

    Phase difference between x = 0 and x = 1 11x

    x sinA

       

    6

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    Phase difference between x = 0 and x = 2x6

    Phase difference between x1 and x2 =  / 3

    8. (D)

    Let v0 is maximum velocity of each particle.

    When particles are on opposite sides of x = 0, let their phase by + & –  (1 1P B 0

    v v v cos 1.2 )

    When they cross each other let the phase be ,2 2P B 0

    v v v cos 1.6

    Phase travelled by Q is  +

    Phase travelled by P is2 2

    (Since 'P' goes to one extreme then comes back to cross Q)since angular frequency is same, phase moved would also be same.

    2 2

    2

    2P 0 0v v cos 1.6 v sin

    2

    2 20v (1.2) (1.6)

    0v 2 m / s

    9. (C)

    From extreme to x = a/2 phase covered = /3

     –A

    A/2

    1/3

    /2

    +A

    60º

    30ºtime taken = T/6

    a / 2 3av

    T / 6 T

    10. (C)

    K.E. at D =1

    max K.E.4

    2 2 2 2 21 1m (A x ) m A2 4

    (CD)

    3 'A 'x

    2

    AE 2A 2R   A 2R BD = 2CD

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     3

    2 R 3R  2

    11. (D)

    Let v A cos( t) 2a A sin( t)

    2 2

    2 2 2 4v a 1

    A A

    2 2 2 2

    2

    1v a A

    Straight line with '–ve' slope

    12. (D)

    Phase moved in T/82 T

    T 8 4

    x a sin( t)

    ax

    2

    13. (C)

    2 2 2x A x x = 1, A = 2

    2 1 4 1

    3

    frequency f3

    2 2

    14. (B)

    K in parallel = 2k K in series = k/2

    15. (C)

    gel. = (g + a) when the elevator accelerates up.

    P

    LT 2 (g a) S MT 2 k 

    TP - downwards, Ts - same

    16. (B)

    Lsin60ºT 2

    g

    60º

    m

                   l

      s   i  n    6

       0   º

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    17. (A)

    y sin( t) 3 cos( t)

    y 2sin t3

    3

    160º

    2

    (max. accl.)A g g

    A

    for maximum acceleration

    y max sin t 13

    g

    t3 2

    t6

    2

    t 6 g

    18. (C)

    IT 2

    mga

     

    23 mR 22mgR 

     3R 

    22g

     3

    2g

    For equivalent length of simple pendulumL

    T 2g

    L = 3

    19. (B)

    2 4U ax bx for equilibrium (mean position)

    3duF 0 2ax 4bx 0dx

    x = 0,2a

    x4b

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    let y be the displacement from mean position3F 2ax 4bx

     puttinga

    x y2b

    3

    a aF 2a y 4b y

    2b 2b

    2a a ay 2a 4b y 2 y2b 2b 2b

    2a a ay 2a 4b 4by 4b.2 .y2b 2b 2b

    2a a2 .y 4b 4b.2 y 02b 2b

    mF 4ay

    4a

    m

    20. (C)

    Particle executes SHM of amplitude 'R'. Initially they col-lide at the centre since their time periods are same

    3

    GM

    2mR mR 3mA

    R A

    3 ( A   new amplitude)

    21. (B)

    Suppose collision occurs at

    Phase covered by 1 is 12

     –A +AO

    Phase covered by 2 is 22 2

     –A

    1

    2

    +A

    90º –

    1 2  (T-same)

    2 2 2

    4

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     phase 13

    2 4

    time takenT 3

    2 4

    3T

    8

    22. (D)

    mT 2

    mT' 2

    T' = T

    23. (A)

    elevator g (g a)

    2

    LT 2

    g a

    1 2T T

    24. (B)

    x Asin( t)

    2 Tx Asin

    T 12

    x = A / 2

    a

    2 2 2

    2 2

    1

    m (A x )K.E. 321P.E. 1m x2

    25. (A)

    1maxv A ( constant)

    2maxv 2A 2v

    26. (C)

    T 2t 1sec

    2 2 k 

    27. (C)

    1y sin3

    2y sin t

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    Phaser  2 2max 1 2 1 2A A A 2A A cos

     1

    1 1 22

    maxA 3

    28. (B)

    1y Asin t

    2y Acos t

    1 2y y 2 Asin t4

    energy 2

    21 m 2A2

      2 2m A

    30. (A)

    Let x Asin( t) 2a A

    3da A cos( t)dt

    for maxda

    cos t 1dt

    at x = 0

    for min da cos t 0dt

    x A

    31. (D)

    centre of mass falls as water comes out, then suddenly amount regainits original position as total of water goes out.

    32. (C)

    1L L

    T 2 2g g

    mgT 2

    Ag

     density of liquid   density of solid 

    x 1Ag L Ag

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    1

    LL

    33. (B)

    atteractive force b/w change and metal plate, geff 

     increases.

    34. (C)

    K.E. = 2 2 21 m (A x )2

        T.E. = 2 21 m A E2

       

    at x = A/2

    K.E. =2

    2 21 Am A2 4

     

    =2 21 3 3Em A

    2 4 4   

    35. (B)

    1 1 2 2 totalk l k l k l

    1

    lk k l

    4

    1k 4k 

    mT 2

    m Tnew T ' 2

    4k 2

    36. (C)

    2

    22

    2

    mgx

    mLmx

    l

    for minimum T is maximum

    22

    22

    222

    2

    mLmx x.2mx

    ld 0 mg

    dx mLmx

    l

     

    L Lx

    12 2 3

    Super position of two SHM's in the same direction will be another SHM if their frequencies are

    equal. Resultant equation of option (B) is1 3y 5sin t tan

    4

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    y 10cos 2 t6

    dy20 sin 2 t

    dt 6

    at t = 1/6

     pv 20 sin3 6

      = –0.628 m/s

    39. (A)

    1y 3sin( t)

    2y 4sin t 3sin( t)2

    using phasor method 

    3

    454

    sin t

    12

    4y 5sin t tan

    3

     phase difference 14

    tan3

       

    40. (A)

    If particle motion starts from extreme

    x Acos( t)

    at t / 6

    x Acos6

     

    maxV V sin( t)

    t / 6

    maxV V sin6

     

    maxVV

    2

    P m v

      maxmV m2E mE

    02 2 2

    41. (C)

    1

    2

    3

    2

    2

    5

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    Relative  = 1 – 2

    time taken to come back in same phaserel

    2t

    1

    2 15t 7.5s

    2 2 5 3

    3 5

    42. (C)

    Let the spring is further extended by y when the cylinder is given small downward push. Then the restoring forceson the spring are,(i) Ky due to elastic properties of spring

    y(ii) upthrust =  yAdg = weight of liquid displaced  Total restoring force = (K  +  Adg)  y   M a (K Adg)y

    Comparing with 2a y,  we get

    2     K Adg

     M   or

      K Adg

     M 

     1

    2 2

    K Adg

     f  M 

    .

    43. (B)

    Maximum tension in the string is at lowest position.

    Therefore2

     Mv

    T Mg L

    To find the velocity v at the lowest point of the path, we apply law of conservation of energyi.e.

    21 (1 cos )2

       Mv Mgh MgL [ , cos ]   h L x h L L

    or  2 2 (1 cos ) v gL

    or 2 (1 cos ) v gL

    2 (1 cos ) T Mg Mg

     x   L

    ah

     Mv L2/

    Mg

    21 2 2sin

    2

     

    T Mg

    2

    1 42

    T Mg [ sin( / 2) / 2 for small amplitudes]

    2[1 ] T Mg

    From figure   a

     L 

    2

    1

    aT Mg

     L  .

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    3. (B)

    2 2v 108 9x 2 2v 9(12 x ) 2 2 2 2v (A x )

      3

    amplitude A 12 acceleration 2a x

      at x = 3  a = –9 × 0.03  = – 0.27 m/s2

    SHM about x = 0

    4. (A, B, C, D)

    The block loses contact with plank when the plank is at its amplitude

    acceleration of block  ba g ( N 0)

    acceleration of plank 2 pa a

    to just leane 2A g

    the contact2

    2

    10

    40 10

      5rad / sec

     2

    T5

    at lowest point of SHM.

    upward acceleration of block = acceleration of plank  g N = 2mg

      = 2A g

    at half waydown acceleration of blockg

    2

     mg 3

     N mg mg2 2

    At mean position, velocity in maximum a = 0  N = mg

    5. (B, C, D)

    2 2v A y

    alsody

    vdt

    v 0  atT

    t2

    2a y max at t T

    F = ma = 0 at3

    t T4

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    atT

    t2

     v = 0  K.E. =0

       P.E. = T.E.

    6. (B, C, D)

    2U 5x 20x

    dvF 10 20

    dx

     = 10(x 2)

      k = 10F = 0 at x = 2 (mean position)

    mT 2

      =0.1

    210

    T

    5

    7. (A, B)

    Ax Asin( t)

    2

    t6

       

    Tt

    12

    00

    vv v cos( t) 3

    2

    00

    aa a sin( t)

    2

    8. (A, B, C)

    2 21 m A2

    = KEmax

    2 2 21 m (A x )2

    = 0.64 × KEmax

    2 2 2A x 0.64A 2 2x 0.36A

    x = 0.6 A = 6 cm

    22 21 AK.E. m A

    2 4

    atA

    x 52

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    3K.E.

    4  

    2 21 3m A2 4

     max P.E.

    9. (B, D)

    x = 3 sin 100 t + 8 cos2 50 t  = 3 sin 100 t + 4 + 4 cos 100 t

    x = 5sin(100t ) 4    SHM

    Amplitude = 5maximum x = 5 + 4 = 9

    10. (D)2a x

      a

    x=0x

    Slope = 2straight line

    11. (C, D)

    xsin( t)

    a

    y1 cos( t )

    a

    22

    2

    x y1 1

    a a

       uniform circle

    x

    dxv a cos( t)

    dt

    4

    dyv a sin( t)dt

    2 2x 4v v v  = constant

    distance  time

    12. (B, C, D)

    2 2

    2 2 2

    x v1

    A A

    ellipse

    2

    a x    straight line2 2

    2 4 2 2

    a v1

    A A

        ellipse

    13. (A, B, C)

    2 2

    2 2 2

    x v1

    A A

    at x = 0 v A 1.0

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    at v = 0 x = A = 2.5

    4

    2T

    4

     = 1.575

    2a A   = 40 cm/s2

    2 2

    v A x 2 2v 4 (2.5) (1)

      = 4 5.25

      = 2 21

    14. (C)

    y A(1 cos2 t)

    y A(2sin t )

    1

    2

    A 2A 1

    1 1 1

    2 2 2

    V A A 2

    V A 2A

    21 1 1

    22 2 2

    a A 2

    a A 1

    15. (B, D)

    Let x Asin( t ) at t = 0

    Ax

    2

    5,

    6 6

    also 0 0v v v cos( t )

    5

    6

    5x Asin t

    6

    x Asin t2 3

     = Acos t3

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    16. (A, B, C, D)

    for equilibriumkx = mgx = 1 cmif released from natural lengthA = 2x = 2 cm

    mf 2 5

    frequency doesn't depend on value of g.

    17. (B)

    The block has v0 at equilibrium

    0

    0

    vA 

    00

    0

    vx sin( t)

    initial phase is zerosince the block is moving is +ve direction.

    18. (A, C)

    Distance of mean position from water level = immersed length= maximum amplitude for equilibrium

    60 a g 3 Lag maximum amplitude = L = immersed length = 20 cm

    mT 2

    3 ag

    19. (A, C)

    Average total energy =2 21 m A

    2

      = maximum K.E.

    root mean square velocity =0v

    2

    mean velocity = 0

    20. (B, C)

    Average 2 21

    KE m A4

    =Average P.E.

      2 f 

    2 2 21KE m A cos ( t)2

    =2 21 1 cos2 tm A

    2 2

     

      f KE= 2 f 

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    EXERCISE 3

    Comprehension - I

    1. (B)

    in experiment Ifrequency = no. of oscillations/sec

    =

    20/ s

    60

    =1

    Hz3

    2. (C)

    frequency is independent of amplitude

    3. (B)

    frequency is also independent of mass

    4. (D) part icle stops at extreme so it drops vertically.

    Comprehension-II

    5. (B)

    Spring cut into 3 equal parts then spring constant of each part becomes 3k   in parallel

    eff 1 2 3k k k k   = 9k 

    mT' 2 9k 

    TT'

    3

    6. (D)

    x cos 60

    x

    x cos 60

    60 60

    x

    kx cos 60 kx cos 60

    6060

    kx

    2netF kx 2kx cos 60

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    =3

    kx2

    eff 

    3k k 

    2

    7. (C)

    eff k in series = k 

    eff k in parallel = 9k 

    Comprehension-III

    8. (D)

    When spring of 2k displaces x, spring of k displaces by 2s (torque balanced about mid  point)

    mid point displacesby 03x

    y2

    02yx3

    netF 2kx k2x

      = 4 kx

    0net

    4k2yF

    3  = 0

    8k y

    3

    energy stored =2

    0

    1 8k (y )

    2 3

      = 204k y3

    9. (A)

    mT 2

    8k 

    3

    10. (B)

    gravity springexternal

    gravity gravity

    w www w

    =

    2 23m3x 1 1k(2x) 2kx2 2 2

    mg3x

    2

    k(2x)2k(x)

    mg

     putting mg = 4kx =1

    2

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    Comprehension-IV

    11. (A)

    Total energy remains constant

    12. (D)

    d Asin( t)

    11 d t sinA

       

    13. (B)

    2 2v A x at x = 0 v   maximumx A v = 0

    Match the Column

    14. (A)  P, R, (B)  R, (C)  P, Q, (D)  P, Q

    m   resF g Ax

    mT 2

    Ag

    (B)  R

    m

    res

    xF (mg vg)

    L

    (C)  P, Q

    Liquid will behave as a point mass(D)  P, Q

    a   area

    x x

    resF g(2x)a

    resF ( g2a)x

    m LT 2 2

    ( g2a) 2g

    15. A  Q, B R, C P, D P

    eff 

    LT 2

    g

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    eff g | g a |

    (A) eff 3g

    g2

    (B) eff g

    g2

    (C)   2

    2eff g g 3g

    (d)eff  22 2

    GMGM GM2g 2 2gR R R 

    2

    16. (A)  R, (B)  S, (C)  P, (D)  P

    17. (A)

     P, Q, (B)

     P, Q, (C)

     S, (D)

     R

    18. A Q, B P, C R, D  S

    y Asin( t)

    v A cos( t)

    (A)21K.E. mv max at t 0

    2

    (B) PE = min at t = 0

    EXERCISE 4

    1. x 0.2cos5 t

    Time period2

    T 0.4s5

    Particle is at x = 0.2 at t = 0

    -A x=0 +A

    t=0

    from x = + A to x = 0it takes 0.1s

    Total distance convered in 0.7 s iss = 7 × A = 7 × 0.2 = 1.4 m

    average speed < v > =Totaldistance

    Totaltime =

    1.4

    0.7= 2m/s

    2. From the given graph

    x x

    comparing with

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    2a x

    frequency1

    f 2 2

    3. T 2 mgd  

    =

    2 22mL mLmL

    12 32

    3L2m

    4

    m 3L4cm

    m

    =17L

    218g

    4. F = –10 x + 2F = – 10 (x – 0.2)k =10

    x=0 mean positionv=0x=-2   0.2

    Amplitudem = 0.1 kg

    10

    0.1 = 10 rad/s Time period =

    2s

    10

    mean position at x = 0.2Amplitude A = + 2 + 0.2 = 2.2m

    equation since particle starts from extremex 0.2 2.2cos t

    x 2.2cos t 0.2

    5. 2U x 4x 3

    (i)dv

    F 2x 4dx

      = –2(x –2) (SHM) equilibrium position at x = + 2

    (ii)

    1

    T 2 22

    (iii) V A

    2 6 A 2 , 2 3m A

    6. Water doesn't roll as the cylinder so it is treated as point mass a.about constant poit

    restoringk(R )R  

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    2I kR 

    2 2 2

    wateras pointmass

    (2MR mR ) kR  

    2 k 

    2M m

    when water becomes ice (neglecting change in volume)ice behaves as solid cylinder 

    2I kR  2 2 232MR mR kR  

    2

    32M m

    2

    7. (a) For small amplitude, the two blocks oscillate together. The angular frequency is

     =k 

    M m

    and so the time period T = 2 M m

    .

    (b) The acceleration of the blocks at displacement x from the mean position is

    a = – 2x =kx

    M m

    The resultant force on the upper block is, therefore,

    ma =mkx

    M m

    This force is provided by the friction of the lower block.

    Hence, the magnitude of the frictional force ism k | x |

    M m

    (c) Maximum force of friction required for simple harmonic motion of the upper block is

    mkA

    M m at the extreme positions. But the maximum frictional force can only be µ mg. Hence

    mkA

    M m = µ mg

    or, A =µ(M m)g

    8. When the elevator is stationary, the spring is stretched to supportthe block. If the extension is x, the tension is kx which should balancethe weight of the block.Thus, x = mg/k. As the cable breaks, the elevator starts falling withacceleration ‘g’. We shall work in the frame of reference of theelevator. Then we have to use a psuedo force mg upward on the block.This force will ‘balance’ the weight. Thus, the block is subjected to a net force kx by the spring

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      I I T P A CE- i an ’ s   Edu.Pvt.Ltd. SIMPLE HARM ON IC MOTION Rg - SHM - 12when it is at a distance x from the position of unstretched spring. Hence, its motion in the elevator is simple harmonic with its mean position corresponding to the unstretched spring. Initially, thespring is stretched by x = mg/k, where the velocity of the block (with respect to the elevator) iszero. Thus, the amplitude of the resulting simple harmonic motion is mg/k.

    9. The situation is shown in figure. The moment of inertia of the disc about the wire is

     =2mr 

    2 =

    2 2(0.200kg)(5.0 10 m)

    2

    = 2.5 × 10 –4 kg - m2.

    The time period is given by

    T = 2 C

    or, C =2

    2

    4

    T

     =

    2 4 2

    2

    4 (2.5 10 kg m )

    (0.20s)

      = 0.25

    2

    2

    kg m

    s

    .

    10. If the string is displaced slightly downward by  x , we can write,the net (restoring)force( 2 )2  x x g

    2  xg

    A B

    Peg

    (5 ) 2   x xg

    or 2

    5

    g x x

    2

    5

    g

    or 2 5

    2  

    T g

     

     

     

    11. When the plank is situated symmetrically on the drums,the reactions on the plank from the drums will be equaland so the force of friction will be equal in magnitude but opposite in direction and hence, the plank will be inequilibrium along vertical as well as in horizontaldirection. Now if the plank is displaced by x to the right, the reactionwill not be equal. For vertical equilibrium of the plank 

     A B

     R R mg …(i)

    R R 

    f A f B

    2L

    mgR BR A

    f A f B

    A Bmgx

    A B

    And for rotational of plank, taking moment about center of mass we have

    ( ) ( )  A B R L x R L x …(ii)Solving Eqns. (i) and (ii), we get

    2

     

     A

     L x R mg

     L

    and 2

       B

     L x R mg

     L

     Now as  f R , so friction at  B will be more than at  A and will bring the plank back, i.e.,

    restoring force here

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    ( ) ( )  B A B Amg

    F f f R R x L

    As the restoring force is linear, the motion will be simple harmonic motion with force constant

      mgk 

     L

    So that 2 2 m L

    T k g

    .

    12. (a) If , the ball does not collide with the wall and it performs full oscillations like a

    simple pendulum.

     period 2  

    g

    (b) If , the ball collides with the wall and rebounds

    with same speed. The motion of ball from A to Q isone part of a simple pendulum.time period of ball 2( )   AQt  .

    AQ

        l

    l   l

    Consider A as the starting point (t = 0)

    Equation of motion is ( ) cos  x t A t 

    ( ) cos ,  x t t 

    ( ) cos ,  x t t   because amplitude  A

    time from A to Q is the time t when x becomes

    cos   t 

    11/ cos 

     AQt t 

    The return path from Q to A will involve the same time interval.

    Hence time period of ball 2   AQt 

    1 12 cos 2 cos

    g

    12 2 cos 

    g g

    13. Suppose that the liquid is displaced slightly from equilibrium so that itslevel rises in one arm of the tube, while it is depressed in the second arm by the same amount,  x .

    h+xh-x

    If the density of the liquid is , then, the total mechanical energy of the

    liquid column is :

    2

    1( ) ( ) .

    2

      dx

     E A h x A h xdt 

    ( ) ( )2 2

    h x h x A h x g A h x g

    2

    2 21 1(2 ) 2 ( )2 2

    dx

     Ah A g h xdt 

    (i)

    After differentiating the total energy and equating it to zero, one finds acceleration

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    The angular frequency of small oscillations, , is:

    2

    2

     A g g

     Ah h(ii)

    14. Suppose that the plank is displaced from its equilibrium position by  x  at time t  , the centre of the

    cylinder is, therefore, displaced by

    2

     x

    the mechanical energy of the system is given by,

    . . E K E  . . E K E  (Plank) + P.E.(spring) + K.E. (cylinder)22

    21 1 1 22 2 2 2

     

    dx d x E m kx m

    dt dt  

    2

    21 1 12 .2 2 2

     

    d xm R

     R dt 

     2

    21 7 1( )2 4 2

    dx

    m kxdt 

    After differentiating the total energy and equating it to zero, one finds acceleration 2   x 

    The angular frequency,

    4

    7

      k 

    m

    15. Suppose that the particle is displaced from its equilibrium position at O , and that its x-coordinate at time t   is given by  x .The total energy of the particle at time t  is given by,,

    2 21

    2

    dx dy E m mgy

    dt dt   (i)

    Differentiating the equation of the curve, we get,

    2 4dx dy

     x a

    dt dt  

    2 22

    2

    11

    2 4 4

     

    dx x mg E m x

    dt a a

     The oscillations are very small, both  x  anddx

    dt  are small. We ignore terms which are

    of higher order than quadratic terms in  x  or,dx

    dt  or, mixed terms.

    2

    21 1

    2 2 2

      dx mg E m x

    dt a(ii)

    After differentiating the total energy and equating it to zero, one finds acceleration = 2   x The angular frequency of small oscillations is, consequently,

    2 . 2

    mg g

    a m a(iii)

    16. At equilibrium the net force on the cylinder is zero in the vertical direction:

    0 net 

    F B W  ,  B  the buoyancy and W  the weight of the cylinder..

    When the cylinder is depressed slightly by  x , the buoyancy increases from  B  to  B B  where:

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     B x Ag

    while the weight W   remains the same.

     the net force, net F B B W  

     B | |  x Ag

    The equation of motion is, therefore,2

    2

    s

    d x Ah x Ag

    dt 

    the minus sign takes into account the fact that  x  and restoring force are in opposite directions.

    2

    2

    s

    gd x  xdt h

    and the angular frequency, , is

    s

    g

    h

    17. Suppose that the rod is displaced by a small angle  as shown in the figure. The totalmechanical energy of the system is given by,

    2 21 (1 cos )3 2

     E m mg21 ( )

    2 k 

    2 21

    3 m

    2 21

    2 2

      mg

    k  (i)

    the angular frequency of small oscillations is,

    A   k kx

    mg

    O

    2

    2

    3 321 23

    mgk 

    k g

    mm

    (ii)

    The condition for the system to be oscillation is,

    3 3

    2

    k g

    m or, 2

    mgk 

    (iii)

    18. Suppose that the block is depressed by  x . The pulley (owing to the constraint) is depressed by

    2

     x. Suppose that the tension in the string are &   T T   on both sides. We can write:

    For block:     mg T mx ...(i)

    For pulley: 0( )2

       x

    T T mg k x x m … (ii)

    The angular acceleration of the pulley, / 2    x

     R… (iii)

    mg

    m

    TT’ x( )

    2

       xT T R I  

     R… (iv)

    From (i), (ii), (iii) and (iv) we get,

    0 2

    53 ( )

    2 2

    m I 

    mg k x x x R

    … (v)

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    The frequency of small oscillation,

    2

    1

    522 2

    k  f 

    m I 

     R

    19. (a) At equilibrium, the net torque on the pulley is zero.

    1 sin m g R mg R … (i)

    or, 1sin  mm

    or,1 1sin

      m

    m… (ii)

    Om  

    m1

    mg

    mg

    T

    O

    (b) If the system is displaced slightly from the equilibrium position, it oscillates. Suppose that the position of the particle is given by the angular variable , at some instant.The total mechanical energy is given by:

    . . . .  E K E P E 

    Om  

    m1

    Owhere,2 2 2 2 2 2

    1

    1 1 1 1. . ( )

    2 2 2 2

    K E MR mR m R

    and, P.E.= loss in P.E. of 1m  + gain in P.E. of m

    1

    ( ) (cos cos ) m gR mgR

    1 ( ) 2 sin sin2 2

    m gR mgR

    1 2 sin sin2 2

    m gR mgR

    2

    1 2 sin 2 cos2 2

      m gR mgR mgR

    where  is defined by the expression : ,  being a small quantity. Since the frequency

    depends only on terms which are quadratic in , we can write,

    2 2 211 1 1 cos ( )

    2 2 2  E M m m R mgR + terms linear in  or, constants.

    After differentiating the total energy and equating it to zero, one finds acceleration = 2 x 

    the angular frequency,2

    1

    cos

    1

    2

    mgR

     M m m R

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    and the frequency,

    1

    1 cos

    12

    2

     

    mg f 

     M m m R

    .

    20. (a) Since the system is in equilibrium, we can write the tension in the string, T   as:2

    1 0 T m r 

    and, 2T m g

    21 0 2 m r m g … (i)

    (b) Suppose that the block 2m  is depressed by  x . The radius of the circle of rotation is now

    given by,

      r r x .

    and the angular speed  is given by,,

    2 21 0 1 ( )   m r m r x

    B’ A’

    (f-x)( +x)l

    C’

    mg

    T

    m g1

    m w (r-x)1 22

    or,2

    0

    2( )

    r x  (ii)

    The free body diagram as well as the geometry of 

    the problem are as shown in the adjacent figure.

    22

    1 12( ) ( )

    d m r x m r x T  

    dt   (iii)

    2

    2 22( )

    d m x m g T  

    dt    (iv)

    The first term on the RHS of the equation (iii) can be rewritten as,

    2 42 1 0

    1 3( )

    ( )

    m r m r x

    r x

    3

    21 0 1

     xm r 

    21 0

    31

      x

    m r r 

    (after binomial expansion and assuming  x r )

    Equation (iii) and (iv) become

    21 1 0

    31

      x

    m x m r T  r 

    2 2 m x T m g .

    Adding,2

    1 2 1 0 2

    3( ) 1

      x

    m m x m r m gr 

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    21 0

    1 2

    3  

      m x x

    m m (v)

    Thus the angular frequency of small oscillations, , is given by,,

    10

    1 2

    3

    m

    m m  (vi)

    EXERCISE 5

    1. (A)

    Potential energy3

    V(x) k | x |

    By the conservation of energy,2 3 31 mv kx ka

    2  (in the region x > 0)

    3 32k v (a x )m

    3 3dx 2k  (a x )dt m

    0t

    3 3 0a

    m dxdt

    2k  a x

    Substituting 2x a sin

    0

    3/2 62

    m 2a sin cos d  

    2k  a 1 sin

    = –t

    0

    2

    1 mt f ( )d  

    k a 

    1T

    a

    2. (D)2x

    xU k(1 e )

    2

    xdU

    F ke .( 2x)dx

      2xk2xe F ( x )  for x   small

    3. (A), (C)

    1 2 3y a sin t; y asin t ; y a sin t4 4

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    1 2 3y y y a 1 2 sin t Energy  (amplitude)2

    Energy 2 2 2a (1 2) a (3 2 2)

    4. (A)

    Effective value of g :g ' g cos

    LT 2 gcos

    5.F Y L

    A L

    YA LF

    L

    Frequency1 YA

    2 LM

    6. (A)

    From 0 A / 2

    Phase covered as6

    From 0 A

    Phase covered as2

    1T6

    22 6T

    2

    1

    T2

    T

    T2 > T

    1

    7. (A)

    x Acos t . At t = 0, the particle is at the extreme position.At the extreme position, the potential energy is maximum and displacement is maximum.

    8. 1 1 2 2 1 2 0m x m x (m m )v t

    12 0

    2

    Amx v t (1 cos t)

    m

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    1 0v v A sin( t)

    When 1 0v v

    sin t 0 cos t 1

    12 0

    2

    Amx v t 2

    m

    1 0x v t 2A

    12 1 0

    2

    mx x 2 2A 1

    m

    l

    1 20

    2

    m mA

    m

    l

    9. (B)

    k 2m

    Maximum acceleration of P

    for P friction provides resortoringF

    2FF m A

    =k kA

    m A2m 2

    10. (B)

    We have,2

    2 2

    2

    d yy kt , 2k 2m / s

    dt

    2122

    T g 2 6

    T g 5

    11. Speed of block at * 2 2y A y

    Height h attaind by block after detachment =

    2 2 *2(A y )

    2g

    Total height attached by the block

    2 2 *2*(A y )H y

    2g

    For H to be maximum, *dH

    0dy

    * 2y g /

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    12. (B), (C)

    If A  B and C 0 then 2 2x Asin t Bcos t Csin t cos t (Not SHM)

    If A = B and C = 2B then x B 2Bsin t cos t B Bsin 2 t (SHM)If A = –B and C = 2B then x Bcos2 t Bsin 2 t (SHM)If A = B and C = 0 then x = A (Not SHM)

    13. (C)

    res L L2 k ( )2 2

    2kL

    2

    L2

    L2

    2

    2

    KL6k 2

    ML M

    12

    14. (D)

    1 2eff 

    1 2

    k k k 

    k k 

    eff 1 1k A k A

    21

    1 2

    k AA

    k k 

    15. (D)

    x Asin( t) T 8s 4

    t , x 1sin t ,3 4

       2

    8 4

    4x 1sin

    4 3

    3x

    2

    Acceleration 2a x 2

    3

    4 2

    223 cm/s

    32

    16. (C)

    At first only left spring (k) is compressed by x

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    2 2 21 1 1kx mv 4ky2 2 2

    xy

    2

    y 1

    x 2

    COMPREHENSION TYPE QUESTIONS

    PASSAGE-1 ( QNO 17 TO 19)

    17. (C)

    Energy should not exceed 0V

    0V E 0

    18. (B)

    2 4 41 mv (A x )2

    4 42v A xm

    x

    4 40

    dx

    A x  =

    t

    0

    2dt

    m

    1 mt

    A

    19. (D)

    for | x | > x0

     potential energy is constantF = 0 a = 0

    PASSAGE 2( Q NO 20 TO 23)

    20. (C)

    0v A At a displacement (x)

    2kxacm

    232kxR mR  2

    cm

    4k x R a

    3M

    4k 

    3m

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    at x = A22kA mg m A

    m4k 2kA mg A

    3m

    3 mgA

    2k 

    03mv A gk 

    21. (D)

    4k 

    3m

    22. (D)

    2net

    m4k F m x x3m

    =4k 

    x3

    23. (A)

    F = QE is a constant forceHence no change in time period 

    24. (A, D)

    In case A2 2

    2m MR mg( / 2)sin Mg sin M3 2ll l l

    In case B2

    2mmg( / 2)sin Mg sin M3

    ll l l

    torque A = torque Bfrequency A < frequency B